Graph y=(2x+3)(x-1)^2

$y = (2 x + 3) {(x - 1)}^{2}$

Step 1

Find the point at $x = - 1$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 2 (- 1) {((- 1) - 1)}^{2} + 3 {((- 1) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $2$ by $- 1$ .

$f (- 1) = - 2 {((- 1) - 1)}^{2} + 3 {((- 1) - 1)}^{2}$

Subtract $1$ from $- 1$ .

$f (- 1) = - 2 {(- 2)}^{2} + 3 {((- 1) - 1)}^{2}$

Multiply $- 2$ by ${(- 2)}^{2}$ by adding the exponents.

Multiply $- 2$ by ${(- 2)}^{2}$ .

Raise $- 2$ to the power of $1$ .

$f (- 1) = (- 2) {(- 2)}^{2} + 3 {((- 1) - 1)}^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- 1) = {(- 2)}^{1 + 2} + 3 {((- 1) - 1)}^{2}$

Add $1$ and $2$ .

$f (- 1) = {(- 2)}^{3} + 3 {((- 1) - 1)}^{2}$

Raise $- 2$ to the power of $3$ .

$f (- 1) = - 8 + 3 {((- 1) - 1)}^{2}$

Subtract $1$ from $- 1$ .

$f (- 1) = - 8 + 3 {(- 2)}^{2}$

Raise $- 2$ to the power of $2$ .

$f (- 1) = - 8 + 3 \cdot 4$

Multiply $3$ by $4$ .

$f (- 1) = - 8 + 12$

Add $- 8$ and $12$ .

$f (- 1) = 4$

The final answer is $4$ .

$4$

Convert $4$ to decimal.

$y = 4$

Step 2

Find the point at $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = 2 (0) {((0) - 1)}^{2} + 3 {((0) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $2$ by $0$ .

$f (0) = 0 {((0) - 1)}^{2} + 3 {((0) - 1)}^{2}$

Subtract $1$ from $0$ .

$f (0) = 0 {(- 1)}^{2} + 3 {((0) - 1)}^{2}$

Raise $- 1$ to the power of $2$ .

$f (0) = 0 \cdot 1 + 3 {((0) - 1)}^{2}$

Multiply $0$ by $1$ .

$f (0) = 0 + 3 {((0) - 1)}^{2}$

Subtract $1$ from $0$ .

$f (0) = 0 + 3 {(- 1)}^{2}$

Raise $- 1$ to the power of $2$ .

$f (0) = 0 + 3 \cdot 1$

Multiply $3$ by $1$ .

$f (0) = 0 + 3$

Add $0$ and $3$ .

$f (0) = 3$

The final answer is $3$ .

$3$

Convert $3$ to decimal.

$y = 3$

Step 3

Find the point at $x = 1$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = 2 (1) {((1) - 1)}^{2} + 3 {((1) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $2$ by $1$ .

$f (1) = 2 {((1) - 1)}^{2} + 3 {((1) - 1)}^{2}$

Subtract $1$ from $1$ .

$f (1) = 2 \cdot 0^{2} + 3 {((1) - 1)}^{2}$

Raising $0$ to any positive power yields $0$ .

$f (1) = 2 \cdot 0 + 3 {((1) - 1)}^{2}$

Multiply $2$ by $0$ .

$f (1) = 0 + 3 {((1) - 1)}^{2}$

Subtract $1$ from $1$ .

$f (1) = 0 + 3 \cdot 0^{2}$

Raising $0$ to any positive power yields $0$ .

$f (1) = 0 + 3 \cdot 0$

Multiply $3$ by $0$ .

$f (1) = 0 + 0$

Add $0$ and $0$ .

$f (1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Step 4

Find the point at $x = 2$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = 2 (2) {((2) - 1)}^{2} + 3 {((2) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $2$ by $2$ .

$f (2) = 4 {((2) - 1)}^{2} + 3 {((2) - 1)}^{2}$

Subtract $1$ from $2$ .

$f (2) = 4 \cdot 1^{2} + 3 {((2) - 1)}^{2}$

One to any power is one.

$f (2) = 4 \cdot 1 + 3 {((2) - 1)}^{2}$

Multiply $4$ by $1$ .

$f (2) = 4 + 3 {((2) - 1)}^{2}$

Subtract $1$ from $2$ .

$f (2) = 4 + 3 \cdot 1^{2}$

One to any power is one.

$f (2) = 4 + 3 \cdot 1$

Multiply $3$ by $1$ .

$f (2) = 4 + 3$

Add $4$ and $3$ .

$f (2) = 7$

The final answer is $7$ .

$7$

Convert $7$ to decimal.

$y = 7$

Step 5

Find the point at $x = 3$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = 2 (3) {((3) - 1)}^{2} + 3 {((3) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $2$ by $3$ .

$f (3) = 6 {((3) - 1)}^{2} + 3 {((3) - 1)}^{2}$

Subtract $1$ from $3$ .

$f (3) = 6 \cdot 2^{2} + 3 {((3) - 1)}^{2}$

Raise $2$ to the power of $2$ .

$f (3) = 6 \cdot 4 + 3 {((3) - 1)}^{2}$

Multiply $6$ by $4$ .

$f (3) = 24 + 3 {((3) - 1)}^{2}$

Subtract $1$ from $3$ .

$f (3) = 24 + 3 \cdot 2^{2}$

Raise $2$ to the power of $2$ .

$f (3) = 24 + 3 \cdot 4$

Multiply $3$ by $4$ .

$f (3) = 24 + 12$

Add $24$ and $12$ .

$f (3) = 36$

The final answer is $36$ .

$36$

Convert $36$ to decimal.

$y = 36$

Step 6

The cubic function can be graphed using the function behavior and the points.

$\begin{array}{c} x & y \\ - 1 & 4 \\ 0 & 3 \\ 1 & 0 \\ 2 & 7 \\ 3 & 36 \end{array}$

Step 7

The cubic function can be graphed using the function behavior and the selected points.

Falls to the left and rises to the right

$\begin{array}{c} x & y \\ - 1 & 4 \\ 0 & 3 \\ 1 & 0 \\ 2 & 7 \\ 3 & 36 \end{array}$

Step 8

Graph y=(2x+3)(x-1)^2

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